Answer:

**Answer:**

Point motion will eventually stops due to **action of g exactly perpendicular...**

**Explanation:**

If ignoring the air resistance, the magnitude of gravitational acceleration is already strong enough to stops the acceleration. As we know that, the spring constant of a bungee spring cord will be** F = -k/x**, where x is the stretched length and k is the spring constant of bungee cord. If **F = ma = w = mg, the g = -m k/x.** Now we can clearly see that the value of g remains constant due to the fluctuating length of the cord as the motion progresses back and forth in **SHM **say from **x1 to x2 and x2 to x1.**

Two wires A and B with circular cross-section are made of the same metal and have equal lengths, but the resistance of wire A is four times greater than that of wire B. What is the ratio of the radius of A to that of B

A train moving west with an initial velocity of 20 m/s accelerates at 4 m/s2 for 10 seconds. During this time, the train moves a distance of meters.

What is the frequency of a photon that has the same momentum as a neutron moving with a speed of 1.90 × 103 m/s?

he magnetic field strength at the north pole of a 2.0-cmcm-diameter, 8-cmcm-long Alnico magnet is 0.10 TT. To produce the same field with a solenoid of the same size, carrying a current of 1.9 AA , how many turns of wire would you need? Express your answer using two significant figures.

(PLEASE HELP ITS DUE SOON ILL MARK BRAINLIEST AND 5 STARS & PLEASE SHOW WORK!!)(And the answer is not 44 I already tried that and it doesn’t start with 4 either)

A train moving west with an initial velocity of 20 m/s accelerates at 4 m/s2 for 10 seconds. During this time, the train moves a distance of meters.

What is the frequency of a photon that has the same momentum as a neutron moving with a speed of 1.90 × 103 m/s?

he magnetic field strength at the north pole of a 2.0-cmcm-diameter, 8-cmcm-long Alnico magnet is 0.10 TT. To produce the same field with a solenoid of the same size, carrying a current of 1.9 AA , how many turns of wire would you need? Express your answer using two significant figures.

(PLEASE HELP ITS DUE SOON ILL MARK BRAINLIEST AND 5 STARS & PLEASE SHOW WORK!!)(And the answer is not 44 I already tried that and it doesn’t start with 4 either)

**Answer:**

Time taken for 1 swing = 3.81 second

**Explanation:**

**Given:**

Time taken for 1 swing = 2.20 Sec

**Find:**

Time taken for 1 swing , when triple the length(T2)

**Computation:**

Time taken for 1 swing = 2π[√l/g]

2.20 = 2π[√l/g].......Eq1

Time taken for 1 swing , when triple the length (3L)

Time taken for 1 swing = 2π[√3l/g].......Eq2

Squaring and dividing the eq(1) by (2)

4.84 / T2² = 1 / 3

T2 = 3.81 second

**Time taken for 1 swing = 3.81 second**

**Answer:**

**Radius at liftoff 8.98 m**

**Explanation:**

At the working altitude;

maximum radius = 24 m

air pressure = 0.030 atm

air temperature = 200 K

At liftoff;

temperature = 349 K

pressure = 1 atm

radius = ?

**First, we assume balloon is spherical in nature,**

**and that the working gas obeys the gas laws.**

from the radius, we can find the volume of the balloon at working atmosphere.

Volume of a sphere =

volume of balloon = x 3.142 x = 57913.34 m^3

using the gas equation,

=

**The subscript 1 indicates the properties of the gas at working altitude, and the subscript 2 indicates properties of the gas at liftoff.**

imputing values, we have

=

0.03 x 57913.34 x 349 = 200V2

V2 = 606352.67/200 = **3031.76 m^3 this is the volume occupied by the gas in the balloon at liftoff.**

from the formula volume of a sphere,

V = = x 3.142 x = 3031.76

4.19 = 3031.76

= 3031.76/4.19

radius r of the balloon on liftoff = = **8.98 m**

**Answer:**

???

**Explanation:**

**Explanation:**

Let us assume that forces acting at point B are as follows.

= 0 ...... (1)

= 0

= 0 .......... (2)

Hence, formula for allowable normal stress of cable is as follows.

T =

= 3925 kip

From equation (1), = -3925

= -3925

= 12877.29 kip

From equation (2), -12877.29 (Cos 60) + W = 0

= 0

W = 6438.64 kip

Thus, we can conclude that **greatest weight of the crate is 6438.64 kip.**

To determine the greatest weight of the crate that can be supported without causing the cable to fail, calculate the normal stress on the cable using σ = F/A, where σ is the normal stress, F is the force on the cable, and A is the cross-sectional area of the cable. Then, compare the calculated normal stress to the allowable normal stress. Consider the angle phi in this calculation by using the equation F = W / sin(ϕ), where F is the **force on the cable**, W is the weight of the crate, and ϕ is the angle with respect to the horizontal.

To determine the greatest weight of the crate that can be supported without causing the cable to fail, we need to calculate the normal stress on the cable and compare it to the allowable normal stress. The normal stress can be calculated using the formula σ = F/A, where σ is the normal stress, F is the force on the cable, and A is the cross-sectional area of the cable. In this case, the force on the cable is equal to the weight of the crate, and the cross-sectional area of the cable can be calculated using the formula A = π*(d/2)^2, where d is the diameter of the cable.

Given that the diameter of the cable is 0.5 in and the allowable normal stress is 21 ksi, we can substitute these values into the equations and solve for the force on the cable:

Calculate the cross-sectional area of the cable: A = π*(0.5/2)^2 = π*(0.25)^2 = 0.1963 in^2

Plug the cross-**sectional **area and the allowable normal stress into the formula for normal stress: σallow = F/A → 21 ksi = F/0.1963 in^2

Solve for the force on the cable: F = 21 ksi * 0.1963 in^2 = 4.1183 ksi*in^2

Therefore, the **greatest **weight of the crate that can be supported without causing the cable to fail is equal to the force on the cable, which is 4.1183 ksi*in^2. However, it's important to note that we also need to consider the angle phi (ϕ) in this calculation. Since the cable goes up from the **load **to point B and then left to a pulley, the weight of the crate will create a vertical component and a horizontal component. To determine the weight of the crate that corresponds to the calculated force on the cable, we need to consider the trigonometric relationship between the force and the weight at an angle. In this case, the angle is 30 degrees, so we can use the equation F = W / sin(ϕ), where F is the force on the cable, W is the weight of the crate, and ϕ is the angle with respect to the horizontal.

Given that ϕ = 30 degrees and F = 4.1183 ksi*in^2, we can substitute these values into the equation and solve for the weight of the crate:

Plug the values into the equation: 4.1183 ksi*in^2 = W / sin(30)

Solve for the weight of the crate: W = 4.1183 ksi*in^2 * sin(30)

Therefore, the greatest weight of the crate that can be supported without causing the cable to fail and at an angle of 30 degrees is equal to the force on the cable, which is 4.1183 ksi*in^2, multiplied by the sine of 30 degrees.

For more such questions on **force on the cable**, click on:

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**Answer:**

E = 9.4 10⁶ N / C, The field goes from the inner cylinder to the outside

**Explanation:**

The best way to work this problem is with Gauss's law

Ф = **E. dA **= qint / ε₀

We must define a Gaussian surface, which takes advantage of the symmetry of the problem. We select a cylinder with the faces perpendicular to the coaxial.

The flow on the faces is zero, since the field goes in the radial direction of the cylinders.

The area of the cylinder is the length of the circle along the length of the cable

dA = 2π dr L

A = 2π r L

They indicate that the distance at which we must calculate the field is

r = 5 R₁

r = 5 1.3

r = 6.5 mm

The radius of the outer shell is

r₂ = 10 R₁

r₂ = 10 1.3

r₂ = 13 mm

r₂ > r

When comparing these two values we see that the field must be calculated between the two housings.

Gauss's law states that the charge is on the outside of the Gaussian surface does not contribute to the field, the charged on the inside of the surface is

λ = q / L

Qint = λ L

Let's replace

E 2π r L = λ L /ε₀

E = 1 / 2piε₀ λ / r

Let's calculate

E = 1 / 2pi 8.85 10⁻¹² 3.4 10-12 / 6.5 10-3

E = 9.4 10⁶ N / C

The field goes from the inner cylinder to the outside

At **constant speed** and *varying position* of the hockey puck, implies a **change** in the** velocity** of the **hockey puck **and **net force **is **acting** on it to keep it in **motion**.

According to **Newton's second law of motion**, the force applied to a an object is directly proportional to the product of mass and acceleration of the object.

F = ma

**Acceleration** is the change in the velocity of an object per change in time of motion.

- At
**constant velocity**, the*acceleration*of an object is*zero*.

- When
*acceleration*of an object is*zero*, the**force**on the object is*zero*. - A
**constant speed**(*magnitude only*) and**change**in the**direction**of the object, implies a**change in velocity**of the object. - at changing velocity, the
*acceleration*on an object is positive, and hence net force acts on the object.

Thus, we can conclude that at **constant speed** and *varying position* of the hockey puck, implies a **change** in the** velocity** of the **hockey puck **and **net force **is **acting** on it to keep it in **motion**.

Learn more here: brainly.com/question/8722829

**Answer:**

**Explanation:**

When the puck is sliding on the ice, there is no force being exerted on the puck to keep it moving forward. Instead, inertia keeps the puck moving forward. Friction between the puck and the ice gradually slows the puck down. You hit a hockey puck and it slides across the ice at nearly a constant speed